# Program to find the Product of diagonal elements of a matrix

Given an N * N matrix, the task is to find the product of the elements of left and right diagonal.

Examples:

```Input: arr[] = 1 2 3 4
5 6 7 8
9 7 4 2
2 2 2 1
Output: 9408
Explanation:
Product of left diagonal = 1 * 4 * 6 * 1 = 24
Product of right diagonal = 4 * 7 * 7 * 2 = 392
Total product = 24 * 392 = 9408

Input: arr[] = 2 1 2 1 2
1 2 1 2 1
2 1 2 1 2
1 2 1 2 1
2 1 2 1 2
Output : 512
Explanation:
Product of left diagonal = 2 * 2 * 2 * 2 * 2 = 32
Product of right diagonal = 2 * 2 * 2 * 2 * 2 = 32
But we have a common element in this case so
Total product = (32 * 32)/2  = 512
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• We need to find out the principal diagonal and seconddary diagonal elements of the matrix. Please refer this article for this [ Program to print the Diagonals of a Matrix ]
• In this method we use one loop i.e. a loop for calculating product of both the principal and secondary diagonals
• Divide the answer by middle element for matrix of odd size

Below is the implementation of the above approach:

## CPP

 `// C++ Program to find the Product ` `// of diagonal elements of a matrix ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the product of diagonals ` `int` `productDiagonals(``int` `arr[][100], ``int` `n) ` `{ ` ` `  `    ``int` `product = 1; ` `    ``// loop for calculating product of both ` `    ``// the principal and secondary diagonals ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// For principal diagonal index of row ` `        ``// is equal to index of column ` `        ``product = product * arr[i][i]; ` ` `  `        ``// For secondary diagonal index ` `        ``// of column is n-(index of row)-1 ` `        ``product = product * arr[i][n - i - 1]; ` `    ``} ` ` `  `    ``// Divide the answer by middle element for ` `    ``// matrix of odd size ` `    ``if` `(n % 2 == 1) { ` `        ``product = product / arr[n / 2][n / 2]; ` `    ``} ` ` `  `    ``return` `product; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr1[100][100] = { { 1, 2, 3, 4 }, ` `                           ``{ 5, 6, 7, 8 }, ` `                           ``{ 9, 7, 4, 2 }, ` `                           ``{ 2, 2, 2, 1 } }; ` `    ``// Function calling ` `    ``cout << productDiagonals(arr1, 4) << endl; ` ` `  `    ``int` `arr2[100][100] = { { 2, 1, 2, 1, 2 }, ` `                           ``{ 1, 2, 1, 2, 1 }, ` `                           ``{ 2, 1, 2, 1, 2 }, ` `                           ``{ 1, 2, 1, 2, 1 }, ` `                           ``{ 2, 1, 2, 1, 2 } }; ` `    ``// Function calling ` `    ``cout << productDiagonals(arr2, 5) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find the Product ` `// of diagonal elements of a matrix ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the product of diagonals ` `static` `int` `productDiagonals(``int` `arr[][], ``int` `n) ` `{ ` ` `  `    ``int` `product = ``1``; ` `    ``// loop for calculating product of both ` `    ``// the principal and secondary diagonals ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// For principal diagonal index of row ` `        ``// is equal to index of column ` `        ``product = product * arr[i][i]; ` ` `  `        ``// For secondary diagonal index ` `        ``// of column is n-(index of row)-1 ` `        ``product = product * arr[i][n - i - ``1``]; ` `    ``} ` ` `  `    ``// Divide the answer by middle element for ` `    ``// matrix of odd size ` `    ``if` `(n % ``2` `== ``1``) ` `    ``{ ` `        ``product = product / arr[n / ``2``][n / ``2``]; ` `    ``} ` ` `  `    ``return` `product; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr1[][] = { { ``1``, ``2``, ``3``, ``4` `}, ` `                        ``{ ``5``, ``6``, ``7``, ``8` `}, ` `                        ``{ ``9``, ``7``, ``4``, ``2` `}, ` `                        ``{ ``2``, ``2``, ``2``, ``1` `} }; ` `    ``// Function calling ` `    ``System.out.print(productDiagonals(arr1, ``4``) + ``"\n"``); ` ` `  `    ``int` `arr2[][] = { { ``2``, ``1``, ``2``, ``1``, ``2` `}, ` `                        ``{ ``1``, ``2``, ``1``, ``2``, ``1` `}, ` `                        ``{ ``2``, ``1``, ``2``, ``1``, ``2` `}, ` `                        ``{ ``1``, ``2``, ``1``, ``2``, ``1` `}, ` `                        ``{ ``2``, ``1``, ``2``, ``1``, ``2` `} }; ` `    ``// Function calling ` `    ``System.out.print(productDiagonals(arr2, ``5``) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 Program to find the Product ` `# of diagonal elements of a matrix ` ` `  `# Function to find the product of diagonals ` `def` `productDiagonals(arr, n): ` ` `  `    ``product ``=` `1``; ` `     `  `    ``# loop for calculating product of both ` `    ``# the principal and secondary diagonals ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# For principal diagonal index of row ` `        ``# is equal to index of column ` `        ``product ``=` `product ``*` `arr[i][i]; ` ` `  `        ``# For secondary diagonal index ` `        ``# of column is n-(index of row)-1 ` `        ``product ``=` `product ``*` `arr[i][n ``-` `i ``-` `1``]; ` `     `  `    ``# Divide the answer by middle element for ` `    ``# matrix of odd size ` `    ``if` `(n ``%` `2` `=``=` `1``): ` `        ``product ``=` `product ``/``/` `arr[n ``/``/` `2``][n ``/``/` `2``]; ` ` `  `    ``return` `product; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr1 ``=` `[[ ``1``, ``2``, ``3``, ``4` `],[ ``5``, ``6``, ``7``, ``8` `], ` `            ``[ ``9``, ``7``, ``4``, ``2` `],[ ``2``, ``2``, ``2``, ``1` `]]; ` ` `  `    ``# Function calling ` `    ``print``(productDiagonals(arr1, ``4``)); ` ` `  `    ``arr2 ``=` `[[ ``2``, ``1``, ``2``, ``1``, ``2` `],[ ``1``, ``2``, ``1``, ``2``, ``1` `], ` `            ``[ ``2``, ``1``, ``2``, ``1``, ``2` `],[ ``1``, ``2``, ``1``, ``2``, ``1` `], ` `            ``[ ``2``, ``1``, ``2``, ``1``, ``2` `]]; ` ` `  `    ``# Function calling ` `    ``print``(productDiagonals(arr2, ``5``)); ` `     `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# Program to find the Product ` `// of diagonal elements of a matrix ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the product of diagonals ` `static` `int` `productDiagonals(``int` `[,]arr, ``int` `n) ` `{ ` ` `  `    ``int` `product = 1; ` `     `  `    ``// loop for calculating product of both ` `    ``// the principal and secondary diagonals ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// For principal diagonal index of row ` `        ``// is equal to index of column ` `        ``product = product * arr[i,i]; ` ` `  `        ``// For secondary diagonal index ` `        ``// of column is n-(index of row)-1 ` `        ``product = product * arr[i,n - i - 1]; ` `    ``} ` ` `  `    ``// Divide the answer by middle element for ` `    ``// matrix of odd size ` `    ``if` `(n % 2 == 1) ` `    ``{ ` `        ``product = product / arr[n / 2,n / 2]; ` `    ``} ` ` `  `    ``return` `product; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[,]arr1 = { { 1, 2, 3, 4 }, ` `                    ``{ 5, 6, 7, 8 }, ` `                    ``{ 9, 7, 4, 2 }, ` `                    ``{ 2, 2, 2, 1 } }; ` `     `  `    ``// Function calling ` `    ``Console.Write(productDiagonals(arr1, 4) + ``"\n"``); ` ` `  `    ``int` `[,]arr2 = { { 2, 1, 2, 1, 2 }, ` `                    ``{ 1, 2, 1, 2, 1 }, ` `                    ``{ 2, 1, 2, 1, 2 }, ` `                    ``{ 1, 2, 1, 2, 1 }, ` `                    ``{ 2, 1, 2, 1, 2 } }; ` `     `  `    ``// Function calling ` `    ``Console.Write(productDiagonals(arr2, 5) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```9408
512
```

Time Complexity: O(N)

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